The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
<h3>
Answer:</h3>
82.11%
<h3>
Explanation:</h3>
We are given;
- Theoretical mass of the product is 137.5 g
- Actual mass of the product is 112.9 g
We are supposed to calculate the percentage yield
- We need to know how percentage yield is calculated;
- To calculate the percentage yield we get the ratio of the actual mass to theoretical mass and express it as a percentage.
Thus;
% yield = (Actual mass ÷ Experimental mass) × 100%
= (112.9 g ÷ 137.5 g) × 100%
= 82.11%
Therefore, the percentage yield of the product is 82.11 %
Answer:
[OH-] = 6.17 *10^-10
Explanation:
Step 1: Data given
pOH = 9.21
Step 2: Calculate [OH-]
pOH = -log [OH-] = 9.21
[OH-] = 10^-9.21
[OH-] = 6.17 *10^-10
Step 3: Check if it's correct
pOH + pH = 14
[H+]*[OH-] = 10^-14
pH = 14 - 9.21 = 4.79
[H+] = 10^-4.79
[H+] = 1.62 *10^-5
6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14
